# Riemann-Lebesgue Theorem

## Theorem

Let $f: \closedint a b \to \R$ be a bounded mapping.

Let $\mu$ be a one-dimensional Lebesgue measure.

Then $f$ is Darboux integrable if and only if the set of all discontinuities of $f$ is a $\mu$-null set.

## Proof

### Necessary Condition

Suppose that $f$ is Darboux integrable.

We need to prove that the set of all discontinuities of $f$ has measure $0$.

Let, for some positive real number $s$:

$A_s = \set {x \in \closedint a b: \map {\omega_f} x > s}$

where

$\map {\omega_f} x = \ds \inf \set {\map {\omega_f} I: I \in N_x}$

where

$\map {\omega_f} I = \ds \sup \set {\size {\map f y - \map f z}: y, z \in I \cap \closedint a b}$
$N_x$ is the set of open subset neighborhoods of $x$

$\map {\omega_f} x$ and $\map {\omega_f} I$ are called the oscillations of $f$ at a point and on a set respectively.

$A_0 = \set {x \in \closedint a b: \map {\omega_f} x > 0}$

is the set of all discontinuities of $f$.

We need to prove that $\map \mu {A_0} = 0$.

If $A_0$ is empty, it follows by Measure of Empty Set is Zero that $\map \mu {A_0} = 0$.

This proves what we wanted for this case.

Now assume that $A_0$ is non-empty.

This means that $\closedint a b$ contains a point $x$ with $\map {\omega_f} x > 0$.

This means, in turn, that there is a strictly positive real number $s$ for which $A_s$ is non-empty.

Choose such a number $s$.

Let $\epsilon > 0$ be given.

By Condition for Darboux Integrability, there exists a subdivision $P = \set {x_0, x_1, x_2, \ldots, x_n}$ of $\closedint a b$ such that:

$\map U P – \map L p < \epsilon$

where:

$\map U P = \ds \sum_{i \mathop = 1}^n \map {M_i} {x_i − x_{i − 1} }$
$\map L P = \ds \sum_{i \mathop = 1}^n \map {m_i} {x_i − x_{i − 1} }$

where:

$M_i = \ds \sup \set {\map f x: x \in \closedint {x_{i - 1} } {x_i} }$
$m_i = \ds \inf \set {\map f x: x \in \closedint {x_{i - 1} } {x_i} }$

Let $J$ be the subset of $\set {1, 2, \ldots, n}$ that satisfies::

$i \in J$ if and only if $\openint {x_{i - 1} } {x_i}$ contains a point in $A_s$.

First, we intend to prove that $\ds \sum_{i \mathop \in J} \paren {x_i − x_{i - 1} } < \dfrac \epsilon s$.

Let $x^{'}_i$ signify a point in $\openint {x_{i - 1} } {x_i} \cap A_s$ whenever $i \in J$.

We have:

 $\ds \epsilon$ $>$ $\ds \map U P – \map L P$ $\ds$ $=$ $\ds \sum_{i \mathop = 1}^n M_i \paren {x_i − x_{i − 1} } - \sum_{i \mathop = 1}^n m_i \paren {x_i − x_{i − 1} }$ Definitions of $\map U P$ and $\map L P$ $\ds$ $=$ $\ds \sum_{i \mathop = 1}^n \paren {M_i - m_i} \paren {x_i − x_{i − 1} }$ $\ds$ $=$ $\ds \sum_{i \mathop = 1}^n \paren {\sup \set {\map f x: x \in \closedint {x_{i - 1} } {x_i} } - \inf \set {\map f x: x \in \closedint {x_{i - 1} } {x_i} } } \paren {x_i − x_{i − 1} }$ Definitions of $M_i$ and $m_i$ $\ds$ $=$ $\ds \sum_{i \mathop = 1}^n \paren {\sup \set {\size {\map f x - \map f y}: x, y \in \closedint {x_{i - 1} } {x_i} } } \paren {x_i − x_{i − 1} }$ Supremum of Absolute Value of Difference equals Difference between Supremum and Infimum $\ds$ $=$ $\ds \sum_{i \mathop = 1}^n \map {\omega_f} {\closedint {x_{i - 1} } {x_i} } \paren {x_i − x_{i − 1} }$ Definition of $\omega_f$ for Oscillation on Real Subset $\ds$ $\ge$ $\ds \sum_{i \mathop \in J} \map {\omega_f} {\closedint {x_{i - 1} } {x_i} } \paren {x_i − x_{i − 1} }$ as $J$ is a subset of $\set {1, 2, \ldots, n}$ $\ds$ $\ge$ $\ds \sum_{i \mathop \in J} \map {\omega_f} {\closedint {x_{i - 1} } {x_i} } \paren {x_i − x_{i − 1} }$ as $\map {\omega_f} {\openint {x_{i - 1} } {x_i} } \le \map {\omega_f} {\closedint {x_{i - 1} } {x_i} }$ by Supremum of Set of Real Numbers is at least Supremum of Subset $\ds$ $\ge$ $\ds \sum_{i \mathop \in J} \inf \set {\map {\omega_f} I: I \in N_{x^{'}_i} } \paren {x_i − x_{i − 1} }$ as $\set {\map {\omega_f} I: I \in N_{x^{'}_i} }$ contains (as an element) $\map {\omega_f} {\openint {x_{i - 1} } {x_i} }$ $\ds$ $=$ $\ds \sum_{i \mathop \in J} \map {\omega_f} {x^{'}_i} \paren {x_i − x_{i − 1} }$ Definition of $\omega_f$ for Oscillation at Point $\ds$ $>$ $\ds \sum_{i \mathop \in J} s \paren {x_i − x_{i − 1} }$ as $\map {\omega_f} {x^{'}_i} > s$ since $x^{'}_i \in A_s$ $\ds$ $=$ $\ds s \sum_{i \mathop \in J} \paren {x_i − x_{i − 1} }$

Therefore:

$\ds \sum_{i \mathop \in J} \paren {x_i − x_{i − 1} } < \dfrac \epsilon s$

Next, we intend to prove that $\map \mu {A_s} = 0$.

We have:

 $\ds A_s$ $\subseteq$ $\ds \paren {\bigcup_{i \mathop \in J} \closedint {x_{i - 1} } {x_i} } \bigcup \bigcup_{i \mathop = 0}^n \set {x_i}$ $\ds$ $\subset$ $\ds \paren {\bigcup_{i \mathop \in J} \closedint {x_{i - 1} } {x_i} } \bigcup \bigcup_{i \mathop = 0}^n \openint {x_i - \frac \epsilon {s \paren {n + 1} } } {x_i + \frac \epsilon {s \paren {n + 1} } }$ as $\set {x_i} \subset \openint {x_i - \dfrac \epsilon {s \paren {n + 1} } } {x_i + \dfrac \epsilon {s \paren {n + 1} } }$

We can now define an upper bound for $\map \mu {A_s}$:

 $\ds \map \mu {A_s}$ $\le$ $\ds \map \mu {\paren {\bigcup_{i \mathop \in J} \openint {x_{i - 1} } {x_i} } \bigcup \bigcup_{i \mathop = 0}^n \openint {x_i - \frac \epsilon {s \paren {n + 1} } } {x_i + \frac \epsilon {s \paren {n + 1} } } }$ Measure is Monotone $\ds$ $\le$ $\ds \map \mu {\bigcup_{i \mathop \in J} \paren {x_{i - 1} } {x_i} } + \map \mu {\bigcup_{i \mathop = 0}^n \openint {x_i - \frac \epsilon {s \paren {n + 1} } } {x_i + \frac \epsilon {s \paren {n + 1} } } }$ Measure is Subadditive $\ds$ $\le$ $\ds \sum_{i \mathop \in J} \map \mu {\openint {x_{i - 1} } {x_i} } + \sum_{i \mathop = 0}^n \map \mu {\openint {x_i - \frac \epsilon {s \paren {n + 1} } } {x_i + \frac \epsilon {s \paren {n + 1} } } }$ Measure is Subadditive $\ds$ $=$ $\ds \sum_{i \mathop \in J} \paren {x_i − x_{i − 1} } + \sum_{i \mathop = 0}^n \frac {2 \epsilon} {s \paren {n + 1} }$ Measure of Interval is Length $\ds$ $<$ $\ds \frac \epsilon s + \sum_{i \mathop = 0}^n \frac {2 \epsilon} {s \paren {n + 1} }$ as $\ds \sum_{i \mathop \in J} \paren {x_i − x_{i − 1} } < \frac \epsilon s$ $\ds$ $=$ $\ds \frac \epsilon s + \frac {2 \epsilon} s$ $\ds$ $=$ $\ds \frac {3 \epsilon} s$

Therefore:

$\map \mu {A_s} < \dfrac {3 \epsilon} s$

Since $\epsilon$ is an arbitrary number greater than zero:

$\map \mu {A_s} = 0$

Next, we intend to prove that $\map \mu {A_0} = 0$.

Consider the set sequence $\sequence {A_{1 / n} }_{n \mathop \in \N_{>0} }$.

This sequence is increasing since $A_{1 / n} \subseteq A_{1 / \paren {n + 1} }$ for every $n \in \N_{>0}$ by the definition of $A_s$.

By the definition of limit of increasing sequence of sets, the limit of $\sequence {A_{1 / n} }_{n \mathop \in \N_{>0}}$ equals $\ds \bigcup_{n \mathop \in \N_{>0} } A_{1 / n}$.

Every set $A_{1 / n}$ contains only points of discontinuity of $f$.

Therefore, $\ds \bigcup_{n \mathop \in \N_{>0} } A_{1 / n}$ is a subset of $A_0$, the set of all discontinuities of $f$.

Also, every point of discontinuity of $f$ belongs to some set $A_{1 / n}$.

Therefore, $A_0$ is a subset of $\ds \bigcup_{n \mathop \in \N_{>0} } A_{1 / n}$.

Accordingly:

$\ds \bigcup_{n \mathop \in \N_{>0} } A_{1 / n} = A_0$

Furthermore, since $\ds \bigcup_{n \mathop \in \N_{>0}} A_{1 / n}$ equals the limit of $\sequence {A_{1 / n} }_{n \mathop \in \N_{>0} }$, $A_0$ equals the limit of $\sequence {A_{1 / n} }_{n \mathop \in \N_{>0} }$.

In other words, $A_{1 / n} \uparrow A_0$ as defined in Limit of Increasing Sequence of Sets.

We have by property (3) of Characterization of Measures:

 $\ds \map \mu {A_0}$ $=$ $\ds \lim_{n \mathop \to \infty} \map \mu {A_{1 / n} }$ as $A_{1 / n} \uparrow A_0$ $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} 0$ as $\map \mu {A_{1 / n} } = 0$ for every $n$ in $\N_{>0}$ since $\map \mu {A_s} = 0$ for every $s > 0$ $\ds$ $=$ $\ds 0$

In other words, $A_0$ is a $\mu$-null set.

Therefore, the set of all discontinuities of $f$ is a $\mu$-null set as $A_0$ is the set of all discontinuities of $f$.

$\Box$

### Sufficient Condition

$\blacksquare$

## Source of Name

This entry was named for Georg Friedrich Bernhard Riemann and Henri Léon Lebesgue.